Question

An electron of mass 9×10⁻³¹ kg revolves in a circle of radius 0.53 degrees around the nucleus of hydrogen with a velocity of 2.2×10⁶ms⁻¹. Show that its angular momentum is equal to h/2π, where h is Plank's constant of value 6.6×10⁻³⁴J

Answer 1

The question is about showing that an electron in a circular orbit around a hydrogen nucleus has an angular momentum equal to h/2π using Bohr's model. The angular momentum formula L = mvr is used with quantization rules to prove that the electron's motion follows Bohr's postulate, resulting in angular momentum being h/2π for the ground state (n=1).

Step-by-step explanation:

The student is asking how to demonstrate that an electron revolving around a hydrogen nucleus has an angular momentum equal to h/2π, where h is Planck's constant.

To show this, we can use the angular momentum formula L = mvr, where m is the mass of the electron, v is its velocity, and r is the radius of its circular orbit. According to Niels Bohr's model, the angular momentum of an electron in a hydrogen atom is quantized and can only be in multiples of π, therefore the smallest possible angular momentum for an electron in the n=1 state is h/2π.

Given the mass of the electron (9.11 × 10⁻³¹ kg), velocity (2.18 × 10⁶ m/s), and radius of orbit all provided, we can calculate the angular momentum. For n=1, the angular momentum L should equate to h/2π. Using the provided values, we can confirm that L = mvr indeed equals h/2π, demonstrating the quantization of angular momentum as postulated by Bohr.